Question

The eccentricity of an ellipse whose centre is at origin is $ \dfrac{1}{2} $ . If one of the directions is $ x = - 4 $ , then the equation of the normal to it at $ (1,\dfrac{3}{2}) $ is.
A. $ 2y - x = 2 $
B. $ 4x - 2y = 1 $
C. $ 4x + 2y = 7 $
D. $ x + 2y = 4 $

Answer 1

Hint: In the equation of an ellipse having x-axis as the major axis, find out the value of a and b. To find out the equation of the normal, we have to first find out the slope of the tangent at any point of the ellipse that will lead us to the slope of normal and will eventually provide us the correct answer.

Complete step-by-step answer:
We know that equation of an ellipse is given as –
 $ \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 $
We are given that eccentricity, $ e = \dfrac{1}{2} $
Also, directrix for an ellipse with x-axis as the major axis is, $ x = \pm \dfrac{a}{e} $
We are given in the question that one of the directions is -4, so we have –
 $
\Rightarrow - 4 = - \dfrac{a}{e} \\
\Rightarrow a = 4e \\
\Rightarrow a = 4 \times \dfrac{1}{2} = 2 \;
  $
For an ellipse, $ {b^2} = {a^2}(1 - {e^2}) $
Putting the values of known quantities in the above formula, we get –
 $
\Rightarrow {b^2} = {(2)^2}[1 - {(\dfrac{1}{2})^2}] \\
\Rightarrow {b^2} = 4(1 - \dfrac{1}{4}) = 4(\dfrac{3}{4}) \\
\Rightarrow {b^2} = 3 \;
  $
Using the calculated values of a and b in the equation of the ellipse, we obtain –
 $ \dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{3} = 1 $
Now we find out the slope of the tangent to the ellipse at any point by differentiating the above equation.
 $
\Rightarrow \dfrac{d}{{dx}}(\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{3}) = \dfrac{d}{{dx}}(1) \\
\Rightarrow \dfrac{{2x}}{4} + \dfrac{{2y}}{3}(\dfrac{{dy}}{{dx}}) = 0 \\
\Rightarrow \dfrac{x}{2} = - \dfrac{{2y}}{3}(\dfrac{{dy}}{{dx}}) \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 3x}}{{4y}} \;
  $
We know that the slope of normal is negative of the reciprocal of the slope of the tangent. So,
 $ slope\,of\,normal = \dfrac{{ - 1}}{{(\dfrac{{ - 3x}}{{4y}})}} = \dfrac{{4y}}{{3x}} $
 $ Slope\,at\,(1,\dfrac{3}{2}) = \dfrac{{4(\dfrac{3}{2})}}{{3(1)}} = 2 $
Equation of a line through one passing point when its slope is known is given as –
 $ y - {y^1} = slope(x - {x^1}) $
Thus, the equation of the normal is,
 $
\Rightarrow y - \dfrac{3}{2} = 2(x - 1) \\
\Rightarrow \dfrac{{2y - 3}}{2} = 2x - 2 \\
\Rightarrow 2y - 3 = 4x - 4 \\
\Rightarrow 4x - 2y = 1 \;
  $
Therefore, the equation of the normal to the ellipse at $ (1,\dfrac{3}{2}) $ is $ 4x - 2y = 1 $ .
So, the correct answer is “Option B”.

Note: A plane curve surrounding two focal points is called an ellipse, it is usually oval shaped. An ellipse is drawn by tracing a point moving in a plane such that the sum of its distance from the two focal points is constant. When the two focal points become the same, the ellipse is converted into a circle.